Answer to Question #184476 in Calculus for Njabulo

Question

 7. (Sections 9.1, 9.2) Consider the R − R function f defined by f (x) = e x ln x. Determine the third order Taylor polynomial T3 (x) of f about the point 1

Answer 1

$The \ Taylor's \ series \ of \ f(x) \ about \ x= a \ is \\ f(x)=f(a)+\frac{f'(a)}{1}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+....(1)\\ Taking\ upto \ 3rd \ degree \ terms \ only \ we \ get \ the \ Taylor's \ polynomial \ of \ degree \ 3\\ \Rightarrow T_3(x)= f(a)+\frac{f'(a)}{1}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3...(2)\\ We \ evalauate \ f(a), \ f'(a)\ , f''(a) \ and \ f'''(a)\ at \ a= 1 \ and \ substitute \ in (2)\\ f(1)=e^1ln(1)=0\because ln(1)=0\\ f'(x)=\frac{d}{dx}(e^xln(x))=e^x\frac{d}{dx}(ln(x))+ln(x)\frac{d}{dx}(e^x)\\ =e^x\frac{1}{x}+ln(x)(e^x)\\ =e^x(\frac{1}{x}+ln(x))\\ \Rightarrow f'(x)=e^x(\frac{1}{x}+ln(x))\\ \Rightarrow f'(1)=e^1(\frac{1}{1}+ln(1))=e \because ln(1)=0\\ f''(x)=\frac{d}{dx}(f'(x))\\ =\frac{d}{dx}(e^x(\frac{1}{x}+ln(x))\\ =e^x \frac{d}{dx}(\frac{1}{x}+ln(x))+(\frac{1}{x}+ln(x))\frac{d}{dx}(e^x)\\ =e^x (-\frac{1}{x^2}+\frac{1}{x})+(\frac{1}{x}+ln(x))(e^x)\\ =e^x (-\frac{1}{x^2}+\frac{2}{x}+ln(x))\\ \Rightarrow f''(x)=e^x (-\frac{1}{x^2}+\frac{2}{x}+ln(x))\\ \Rightarrow f''(1)=e^1 (-\frac{1}{1^2}+\frac{2}{1}+ln(1))=e\\ f'''(x)=\frac{d}{dx}(f''(x))\\ =\frac{d}{dx}(e^x (-\frac{1}{x^2}+\frac{2}{x}+ln(x)))\\ =e^x\frac{d}{dx}(-\frac{1}{x^2}+\frac{2}{x}+ln(x))+(-\frac{1}{x^2}+\frac{2}{x}+ln(x))\frac{d}{dx}(e^x)\\ =e^x(\frac{2}{x^3}-\frac{2}{x^2}+\frac{1}{x})+(-\frac{1}{x^2}+\frac{2}{x}+ln(x))(e^x)\\ =e^x(\frac{2}{x^3}-\frac{3}{x^2}+\frac{3}{x}+ln(x))\\ \Rightarrow f'''(x)=e^x(\frac{2}{x^3}-\frac{3}{x^2}+\frac{3}{x}+ln(x))\\ \Rightarrow f'''(1)=e^1(\frac{2}{1^3}-\frac{3}{1^2}+\frac{3}{1}+ln(1))=2e\\ \Rightarrow f'''(1)=2e\\ Substituting \ the \ values \ of \ f(1),\ f'(1),\ f''(1) \ and \ f'''(1) \ in \ (2), \ we \ get \\ T_3(x)= f(1)+\frac{f'(1)}{1}(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3\\ =0+e(x-1)+\frac{e}{2!}(x-1)^2+\frac{2e}{3!}(x-1)^3\\ =e(x-1)+\frac{e}{2}(x-1)^2+\frac{e}{3}(x-1)^3\\ \therefore, T_3(x)=e((x-1)+\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3)$